AP Physics 1  ·  Unit 4: Linear Momentum  ·  Lesson 4.3

Deep Dive: Conservation of Linear Momentum

🔬 Deep Dive
This is your textbook for this topic. Take your time. Read it more than once.
Lesson 4.3: Conservation of Linear Momentum
4.3.A.1Concept

Isolated Systems

Conservation of momentum applies to isolated systems — systems on which there is no net external force. Understanding what makes a system isolated is the first step in every conservation problem.

Internal forces

Forces that objects inside the system exert on each other.

Collision contact force between two carts. Explosive force between two fragments. Spring force between two connected objects.
Change individual momenta but always come in Newton's Third Law pairs — net effect on total momentum is zero.
External forces

Forces that objects outside the system exert on objects inside it.

Friction from the floor on a sliding object. Gravity from Earth on a projectile. Normal force from a wall.
Change the total momentum of the system. When external forces are present, momentum is NOT conserved.
💡Many real situations are approximately isolated even when external forces exist. During a very brief collision, the internal collision forces are enormous compared to external forces like friction. We treat the system as isolated for the duration of the interaction — this is the impulse approximation.
4.3.A.2MathConcept

Conservation of Linear Momentum

In an isolated system, the total linear momentum before any interaction equals the total linear momentum after:

Sp_i = Sp_f

Expanded for a two-object system:

m₁v₁i + m₂v₂i = m₁v₁f + m₂v₂f

This equation has four velocity terms and two known masses. In a typical problem you know three of the four velocities and solve for the fourth. Always assign a sign convention before substituting.

🔑Strategy for every conservation problem:(1) Define your system and check for net external forces. (2) Set a positive direction. (3) Write Sp_i = Sp_f with every object on both sides. (4) Substitute known values with signs. (5) Solve for the unknown.
ExampleWorked Example — Two Carts Collide

Cart A (4 kg) moves right at 5 m/s and hits stationary Cart B (6 kg). After the collision Cart A moves right at 1 m/s. Find Cart B's final velocity.

4.3.A.3Concept

Why Momentum Is Conserved — Newton's Third Law

Conservation of momentum is not an assumption — it follows directly from Newton's Third Law. Here is the proof:

// Newton's Third Law:
F_12 = -F_21   (A on B = -(B on A))
// Both act for same time Dt:
F_12*Dt = -F_21*Dt
J_on2 = -J_on1   (impulses cancel)
Dp_2 = -Dp_1   (J = Dp)
Dp_1 + Dp_2 = 0 → Sp is constant ✓
🔑Every internal force has an equal and opposite reaction force. The two impulses cancel exactly. Therefore internal forces can never change total system momentum. Only external forces can. This is why conservation works.
4.3.A.4Math

1D Applications — Collisions and Explosions

Conservation of momentum applies to all interactions in isolated systems: collisions where objects bounce or stick, and explosions where a single object breaks into parts. The equation is the same — only the setup differs.

Set masses and initial velocities. The simulator shows momentum before the collision and calculates what happens if they stick together — verifying total momentum is conserved.

Mass 1 (m₁)3 kg
Mass 2 (m₂)2 kg
Velocity 1 (v₁)+4 m/s
Velocity 2 (v₂)-1 m/s
BEFORE collision
p₁ = m₁v₁
+12.0 kg·m/s
p₂ = m₂v₂
-2.0 kg·m/s
Σp (total)
+10.0 kg·m/s
AFTER (perfectly inelastic — they stick)
Σp (total)
+10.0 kg·m/s
v_f = +2.00m/s  (both objects at same speed)
Σp_before = 10.0kg·m/s  =  Σp_after = 10.0kg·m/s  ✓

Try setting v₁ and v₂ equal and opposite with equal masses — Σp = 0 and both objects stop. Total momentum was zero before and zero after.

ExampleGuided Example — Explosion from Rest

A 6 kg object at rest on a frictionless surface explodes into two pieces. Piece 1 (2 kg) flies left at 9 m/s. Find the velocity of Piece 2 (4 kg).

Step 1 — Check isolation
Frictionless surface — no external horizontal force. System is isolated. Conservation applies.
4.3.A.5Math

Center-of-Mass Velocity

The center-of-mass velocityof a system is the velocity of the point that represents the average position of mass in the system, weighted by each object's mass:

v_cm = Sp / Sm = (m₁v₁ + m₂v₂) / (m₁ + m₂)

In an isolated system, v_cm is constant — it never changes, no matter what interactions occur internally. This is a direct consequence of conservation of momentum: Sp is constant and Sm is constant, so their ratio is constant.

💡v_cm is also the final velocity in a perfectly inelastic collision (when objects stick together). After they stick, the combined object moves at exactly v_cm of the original system. This gives you a fast way to check your perfectly inelastic answers.
ExampleWorked Example — Center-of-Mass Velocity

Object A (3 kg) moves right at 6 m/s. Object B (5 kg) moves left at 2 m/s. Find v_cm of the system before any collision.

← Back to Lesson 4.3Next: Lesson 4.4 →Elastic and Inelastic Collisions — the season finale.
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